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sasho [114]
4 years ago
5

A 1400-kg car is traveling with a speed of 17.7 m/s. What is the magnitude of the horizontal net force that is required to bring

the car to a halt in a distance of 33.1 m?
Physics
1 answer:
AleksandrR [38]4 years ago
5 0

Answer:

The magnitude of the horizontal net force is 13244 N.

Explanation:

Given that,

Mass of car = 1400 kg

Speed = 17.7 m/s

Distance = 33.1 m

We need to calculate the acceleration

Using equation of motion

v^2-u^2=2as

Where, u = initial velocity

v = final velocity

s = distance

Put the value in the equation

0-(17.7)^2=2a\times 33.1

a=\dfrac{-(17.7)^2}{33.1}

a=-9.46\ m/s^2

Negative sign shows the deceleration.

We need to calculate the net force

Using newton's formula

F = ma

F =1400\times(-9.46)

F=-13244\ N

Negative sign shows the force is opposite the direction of the motion.

The magnitude of the force is

|F| =13244\ N

Hence,  The magnitude of the horizontal net force is 13244 N.

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You have a stopped pipe of adjustable length close to a taut 62.0 cmcm, 7.25 gg wire under a tension of 4710 NN. You want to adj
Oksana_A [137]

Answer:

6 cm long

Explanation:

F = 4110N

Vo(speed of sound) = 344m/s

Mass = 7.25g = 0.00725kg

L = 62.0cm = 0.62m

Speed of a wave in string is

V = √(F / μ)

V = speed of the wave

F = force of tension acting on the string

μ = mass per unit density

F(n) = n (v / 2L)

L = string length

μ = mass / length

μ = 0.00725 / 0.62

μ = 0.0116 ≅ 0.0117kg/m

V = √(F / μ)

V = √(4110 / 0.0117)

v = 592.69m/s

Second overtone n = 3 since it's the third harmonic

F(n) = n * (v / 2L)

F₃ = 3 * [592.69 / (2 * 0.62)

F₃ = 1778.07 / 1.24 = 1433.927Hz

The frequency for standing wave in a stopped pipe

f = n (v / 4L)

Since it's the first fundamental, n = 1

1433.93 = 344 / 4L

4L = 344 / 1433.93

4L = 0.2399

L = 0.0599

L = 0.06cm

L = 6cm

The pipe should be 6 cm long

6 0
3 years ago
Please help im failing!!!
natta225 [31]
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5 0
3 years ago
A charged particle moving through a magnetic field at right angles to the field with a speed of 24.7 m/s experiences a magnetic
Sunny_sXe [5.5K]

Answer:

Explanation:

let the charge is q. velocity, v = 24.7 m/s

magnetic force, F = 2.38 x 10^-4 N

Let the magnetic field is B.

Velocity, v' = 5.64 m/s

angle, θ = 21.2°

The force experienced by a charged particle placed in a magnetic field is given by

F = q x v x B x Sinθ

in first case

2.38 x 10^-4 = q x 24.7 x B x Sin 90 .... (1)

in second case

F = q x 5.64 x B x Sin 21.2°      .... (2)

Divide equation (2) by equation (1), we get

\frac{F}{2.38\times 10^{-4}}=\frac{5.64\times Sin 21.2}{24.7\times Sin 90}

F = 1.97 x 10^-5 N

7 0
4 years ago
Consider a circuit with two resistors in parallel R_1 = 10 ohm and R_2 = 5 ohm.A) Determine the total resistance of the circuit.
Sophie [7]

Answer:

Explanation:

Given

R_1=10 \Omega

R_2=5 \Omega

when resistance in Parallel

\frac{1}{R_{p}}=\frac{1}{R_1}+\frac{1}{R_2}

R_p=\frac{R_1R_2}{R_1+R_2}

R_p=\frac{10}{3}

Suppose V is voltage of battery

Total Current i=\frac{3V}{10}

Since Circuit is Parallel therefore Voltage across both resistor is same

V=i_1R_1=i_2R_2

and i_1+i_2=i

i_1+i_1\cdot \frac{R_1}{R_2}=i

i_1(1+\frac{10}{5})=\frac{3V}{10}

i_1=\frac{V}{10}

i_2=\frac{2V}{10}

(b) When Circuit is in series

R_s=R_1+R_2

R_s=10+5=15 \Omega

since circuit is in Series therefore current is same in both resistor

Current i=\frac{V}{15} A

Voltage drop across R_1=i\times R_1

V_1=\frac{V}{15}\times 10=\frac{2V}{3}

V_2=\frac{V}{15}\times 5=\frac{V}{3}              

8 0
3 years ago
What is the potential energy of a spring that is stretched 0.15 m from equilibrium and has a spring constant of 0.55 N/m?
yan [13]

Explanation:

EE = ½ kx²

EE = ½ (0.55 N/m) (0.15 m)²

EE = 0.62 J

8 0
3 years ago
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